To find the derivative of the function

$y = \frac{\arcsin(x - 1)}{2} + \ln(4 - x)$

we’ll differentiate each term separately.

Step 1: Differentiate $\frac{\arcsin(x - 1)}{2}$

Let $u = x - 1$. Then, $\arcsin(x - 1)$ becomes $\arcsin(u)$.

Using the chain rule:

1. Differentiate $\arcsin(u)$ with respect to $u$: $\frac{d}{du} \arcsin(u) = \frac{1}{\sqrt{1 - u^2}}$
2. Differentiate $u = x - 1$ with respect to $x$: $\frac{du}{dx} = 1$

So, by the chain rule: $\frac{d}{dx} \arcsin(x - 1) = \frac{1}{\sqrt{1 - (x - 1)^2}} = \frac{1}{\sqrt{2x - x^2 - 1}}$

Now, because this term is divided by 2, the derivative of $\frac{\arcsin(x - 1)}{2}$ with respect to $x$ is: $\frac{1}{2} \cdot \frac{1}{\sqrt{2x - x^2 - 1}}$

Step 2: Differentiate $\ln(4 - x)$

Let $v = 4 - x$.

Using the chain rule:

1. Differentiate $\ln(v)$ with respect to $v$: $\frac{d}{dv} \ln(v) = \frac{1}{v}$
2. Differentiate $v = 4 - x$ with respect to $x$: $\frac{dv}{dx} = -1$

So, by the chain rule: $\frac{d}{dx} \ln(4 - x) = \frac{-1}{4 - x}$

Combine the Results

Now we combine the derivatives of both terms:

$y' = \frac{1}{2} \cdot \frac{1}{\sqrt{1 - (x - 1)^2}} - \frac{1}{4 - x}$

Thus, the derivative of the function $y = \frac{\arcsin(x - 1)}{2} + \ln(4 - x)$ is:

$y' = \frac{1}{2\sqrt{1 - (x - 1)^2}} - \frac{1}{4 - x}$

Would you like further details or have questions on any steps?

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Related Questions

1. How can we interpret the derivative in terms of the behavior of the function?
2. What are the restrictions on $x$ for the derivative to be valid?
3. How does the term $\ln(4 - x)$ affect the overall slope as $x$ approaches 4?
4. Can this derivative be simplified further under certain conditions?
5. How does the addition of two different function types (arcsin and logarithm) impact the function's overall behavior?

Tip: When differentiating functions with nested expressions, break down each step with chain rule applications to ensure accuracy.